- Email: [email protected]
On the structure of subrecursive degrees
JOURNAL OF COMPUTER AND SYSTEM SCIENCES 4, 452--464
(1970)
On the Structure of Subrecursive Degrees *t SANAT K. Basu:
Carnegie-Mellon University, Department of Computer Sciences, Pittsburgh, Penn. 15213 Received April 30, 1969
Subrecursive degrees are partitions of computable (recursive) functions generated by strong reducibility orderings. Such reducibilities can be naturally characterized in terms of closure operations. Closure operations corresponding to standard reducibilities such as "primitive recursive," etc., are computation time closed. It is shown that if the closure operation defining a strong reducibility satisfies certain axioms, then the partial ordering of the subrecursive degrees contains dense chains.
PRELIMINARIES Let X denote the nonnegative integers and ~ , ( ~ ) be the class of partial-recursive (recursive) functions of n variables. We shall use the standard indexing of partial recursive functions, satisfying the universal function theorem and the s-m-n theorem [1]. Let ~i denote the partial function computed by the i-th Turing Machine in the standard indexing. @i will denote the step counting function corresponding to ~i 9 @i(x) will be interpreted as the number of steps used by the i-th T.M. in computing ~i(x) if ~i(x) converges. ~i(x) is undefined otherwise. If ~i is a recursive function, then so is ~ i Axiomatic characterization of the step counting functions is given by Blum [1]. DEFINITION. A computable function f ( = ~i) is honest if the predicate [y = f(x)] is elementary in x and y. Equivalently, ~i is honest if there is an elementary function v(x, y) such that ~i(x) < v(x, $i(x)) for almost all x. A mapping F from functions of one variable to functions of one variable is called an operator. For any function f, the value of F(f) at x will be written as F(f, x). If 9 An earlier version of these results were presented at the ACM Symposium on Theory of Computing, Los Angeles, May 1969. 9 This work was supported by the Advanced Research Projects Agency of the Office of the Secretary of Defense (F 44620-67-C-0058) and is monitored by the Air Force Office of Scientific Research. 9 Present address: Computer Centre, Department of Electrical Engineering, Indian Institute of Technology, Kanpur, India. 452
SUBRECURSIVE DEGREES
453
f: ~V"~ ~/" let f(x) be an effective encoding o f f restricted to the domain {0, 1,..., x). Let .A/'~ be the class of total functions from ./V into ~ and let ~- be the class of all partial functions of one variable. DEFINITION. An operator F: ..UW--~ JV W is continuous if there is a function f of two arguments (called the associate of F) such that for every g e ,A/W, x e dff,
F(g, x) = f(~(z), x) -- l, where z is the least integer y such that
f(~(y), x) ~ O. DEFINITION. An operator F: .5"w--+ X W is general recursive if it is continuous and has an associate in ~2 9 I f f e J V W is a total function, then ft-I for n e~V" denotes the restriction o f f to the domain {0, 1,..., n}. LetF, g, and z be as above. Iff[zl ----gtZl thenF(g, x) = F(f, x), and further, z is the smallest integer for which this is true. Thus, if F is general recursive, then for all f ~ ./V"~r and all x (p.z ~ ~C) (Vg ~ ./VW) [f[~l = g[~] ~ F ( f , x) = F(g, x)] is defined and given the function f, can be computed using the recursive associate function of F.
I NTRODUCTION
The notion of primitive recursive degrees was first introduced by Kleene [3] and subsequently studied by Axt [2]. It was shown that the primitive recursive degrees in a given Turing degree form an upper semilattice and that there exist incomparable primitive recursive degrees in any given Turing degree. Restricting our attention to the recursive Turing degree, the primitive recursive degrees are still of interest, since they embody an interesting relation between computable functions, viz., the mutual ease of computability. If f and g are any two recursive functions, we say that f is primitive recursive in g if an oracle Turing Machine with a g oracle can computer in a number of steps bounded by a primitive recursive function. We say thatf and g belong to the same primitive recursive degree i f f is primitive recursive in g and g is primitive recursive in f. The choice of "primitive recursive" in the above discussion is clearly arbitrary. We could equally well consider "elementary degrees," "two-recursive degrees," etc. In general such reducibility orderings can be characterized by closure operations on functions such that a function f is reducible to a function g i f f is contained in the closure of g. All such closure operations must be effective if the induced reducibility ordering is to be stronger than Turing reducibility. Further they also have
454
BASU
the property of computation time closure, i.e., a function is in the class i f f f a (relative) step-counting function of it is also in the class. We have called the degrees of computable functions generated by reducibility ordering of the kind considered above "subrecursive degrees." Axt [2] has shown, that there exist denumerably many incomparble subrecursive degrees. Essentially similar results were obtained in Ref. [6] using only the boundedness property of the corresponding classes. In this paper we show that there exist dense chains in the reducibility ordering of the subrecursive degrees, using an axiomatization of the closure operations. In accordance with our preoccupation with computable functions, we have considered a closure to be a mapping, F, from sets of functions to sets of functions cO
satisfying the following axioms. Vx stands for almost all x. L e t / ' : 2 ~ X -~ 2 ~r'*" be a partial function./" is a complexity closure if there exists a general recursive operator F such that 1. f ~ ~1 ~ {f} ~ dom F
and
F{f} C ~x ;
2. f e r(g} ~ r(f} C r(g); o~
3. f e ~x =- (Vg e F(f})(Vx)[F(f, x) > g(x)]; 4. $j e ~x
and
f > r
~ $ i e F{f}.
We say that f is reducible to g i f f 6 F{g}. Axioms 1 and 2 are standard properties of closure. Axiom 3 provides an upper bound on the functions in the class and to that extent a closure satisfying Axiom 3 is constructive. Axiom 4 is weaker than the property of computation-time closure. In fact, if ~ is a closure operation such that q~(f) is computation-time closed a n d r e ~(f), then ~ satisfies Axiom 4. Clearly, therefore, standard closure operations such as "primitive recursive in," "elementary in" satisfy the axioms. Whether these are the weakest set of axioms under which the results stated in this paper can be derived is not known.
1. Well-Behaved Operators Consider a complexity closure F that satisfies Axioms 1--4. We now make precise the reducibility ordering induced b y / ' . DEFIr~ITION 1.1. A function f is F-reducible to a function g(f <~rg) if r E F(g). f is strictly F-reducible to g if f ~ r g and f ~ r g. By Axiom 2, ~
SUBRECURSIVE DEGREES
455
DEFINITION 1.2. Let G be any general recursive operator. The iterate of G is defined as
Go(f, x) = f(x), Ca(f, x) = C(f, x), Gn+l(f~ x) = G(Gr~(f), x). DEFINITION 1.3. Let G be a general recursive operator and f, g recursive functions. Then f G-dominates g(f >v g) if there exists a monotone, unbounded oo
recursive function r such that (Vx)[f(x) > Gr[z](g, x)]. DEFINITION 1.4. A general recursive operator G is well-behaved if the following hold for allf, g ~ J / ' ~ :
1. G(f) > f; 2. (Vx)[f(x) > g(x)] => (Vx)[G(f, x) > G(g, x)]; 3. (Vf e JV"z)(Vx, y)[x > y ~ G(f, x) > G(f,y)]; 4. (Vfe Mr~)(Vx 9 JV')(3 g 9 JV'~)[f[ x] -- g[=] and G(f, x) =/= G(g, x)]. Thus if is well-behaved, it preserves the majorising relation between functions. Also, G(f) is monotone for each total function f. (4) asserts that the (initial) segment o f f needed to compute G(f, x) is greater than x. Using induction it is easy to prove the following. LEMMA1.1. l f ageneral recursive operator G is well-behaved then so is Gn for all n ~ ~/*. We next prove the following, LEMMA 1.2. Let G be any given general recursive operator. Then we can effectively find a well-behaved general recursive operator F such that for all f ~ ~, F(f) > G(f)
Proof. Consider the step-counting function q~i of a Turing Machine computing a recursive functionf. Then r ~ x and r ~ f ( x ) , for firstly, the Turning Machine computing f(x) must read through the input before the computation could terminate, and secondly, the largest number the machine could write (in unary notation) is equal to the number of steps spent in computation. Suppose further that g 6 JV ~ is a total function and let ~r be a g oracle Turning Machine (OTM) with ~-(g) = ((x, g(x)) [ x ~ JV'} as the oracle, computing g. Then J/~g) needs at least g(x) steps to compute g(x) by sucessively questioning the oraele as to whether (x, O) ~ r(g), (x, 1) E -r(g),..., ( x, g(x)) E "r(g). Sinceg is total, this computation always halts. In terms of a relative step-counting function if r = g then ~b~~ g(x). For f E JV"~, let 7t(f, x) = {g ~ JV"W {g <~ Az max (f(z), x)}. Let gtq
456
BASU
be the restriction of g c ~g'~r to the domain {0, 1,..., n}. Let T(f, x) = {g[-l[ n c ~ - x , g e W(f, x)} U {;~}. Partially ordered under set inclusion, T(f, x) forms a tree with { ~} as the root and with finite branching at each vertex. Since G is general recursive and hence continuous, computation of G(g, x) for any g E ~(f, x) needs only a finite initial segment g. By K6nig's Lemma [5], for any x ~ JV', the set {G(g, x)l g ~ 7t(f, x)} is finite and is determined by a finite subset of T(f, x). In fact, S{G(g, x)i g E ~ ( f , x)}, as a function of x, is recursive in f. For f e ~ , let F(f, x) = q~1~(x), where ~b~11 -- h and
h(O) = [I{G(g, O) I g ~ 7t(f, 0)}] -t- f(0), h(x + 1) = [l{G(g,x + 1 ) [ g e T - ' ( f , x + 1)] + f ( x + 1) + F ( f , x ) . F is general recursive and maps recursive functions into the set of (recursive) stepcounting functions. Properties (1), (3), and (4) of well-behaved operators follow from the definition of F. We show that property (2) holds. Let f, k ~ .ArX, and (Vx)f(x) > k(x). Let h' be the function obtained by replacing f by k in the definition of h. Now (Vx)f(x) > k(x) ~ (Vx) tp(f, x) D ~(k, x) and also k 6 W(f, x). It follows then that F(f, 0) > F(k, 0). Assume as inductive hypothesis that F(f, x) > F(k, x). Then since the number of steps needed by a Turing Machine to compute a step-counting function is equal to the value of the function, and f(x) > k(x) => qb~"(x) > q~'(x) where ~ " - - f and ~k, = k, it is easily seen that F(f, x q- 1) > F(k, x + 1), which proves the lemma.
2. Dense Chain @Functions Suppose F is a given complexity closure with a general recursive operator G. Then using Lemma 1.2 above, we can effectively find a well-behaved general recursive operator F that also satisfies Axiom 3. Using this operator, we next prove the following: THEOREM 2.1. Let F be a well-behaved general recursive operator and let fx,f2 be strictly increasing recursive functions such that fl > ~fz. Then there exists a monotone, unbounded recursive function m such that
A >~ax[F'~'x'(A, x)] >~f~. Proof. fl > Ff2 implies that there exists a monotone, unbounded recursive function r' such thatfl(x ) >Fr'{x}(f2 , x)for almost all x. Since F ~ is well-behaved for all n ~ ~/', F"(f2) is also an increasing function of x. Let r(x) be a monotone unbounded recursive function such that r'(x) > r(x) for almost all x. Then fl(x) ), Fr{~l(f2, x) for almost all x. Define k~ = r(x) if r(x) is even, r(x) -- 1 otherwise. We shall construct m in stages, n~' denotes the segment o f f defined at the end of stage x.
457
SUBRECURSIVE DEGREES
Stage O. Set m(0) = 0, n o = 0. Stage x.
Compute kx 9 Let nx be the smallest integer z such that (Vfm,A/'~)[f ['] < [Fk'/e(f2)]tz] ~ F~'/2(f, x) < Fk'(f2 , x)] "-'.
(1)
We compute n~ as follows: Let hx :Fk'/~(f2). Then h , ~ , and let T(h,) : { f e J f f ' ~ I f < h~}. Define T(hx) = {fin] I n ~ h / ' , f e 7t(h,)} u {~}. By an argument parallel to that used in the proof of Lemma 1.2, there is a finite subset of T(h~) that defines the set {Fk,/2(f, x ) l y e T(h,)}. This finite subset corresponds to a finite tree obtained from T(h,) by ordering it under set inclusion. Let n, be the length of the longest path in this subtree. SinceF is general recursive and h~ is recursive, n~ is effectively computable. Then for all g ~ JV"~', gtn,] < [Fk:12(f2)][,,,] ~ (3 g' E Jff~)[g' E tP(h~) and g'[",] :
gin=I]. Also, g ' e 7J(h~)=~ g ' <
h~ and since F is well-behaved, Fk,/2(g ')
Fk'(f2). Further,
by
construction
of
n~, n~ ~ (~z c ~U)(Vf c jCr~r)[f[z] _-- g,[~] =v
tk,/2(f, x) = p,/2(g,, x)]. But g[n,] = g,[n,], which implies that Fk,/2(g, x) = Fk,/2(g ', x). Then for all g e .A/"~, gln,] .< [Fk:/~(A)][n,] ~ F~:12(g, x) =- Fk:/2(g ', x) .< Fk,12(F~,/2(f~ , x) = F~,(f2 , x). Therefore n~ has the property asserted in (1). Let n~' = max[n~, n~_i], n o' - 0. Set re(y) = k~/2 for n~_ 1 ~< y ~< n~. We next show #
1.
#
n~' is an unbounded function of x.
By construction of n~, for any g ~ 7x(P:/2(f2)), n, >~ (t~z ~ ./V')(Vf~ .,e'~)[f[:] = g[:l =~ p : / 2 ( f , x) = F~:12(g, x)]. F ~,/~ is well-behaved for all x, and therefore by condition 4 of Definition 1.4, n~ ~ x. Thus n~' is unbounded and m gets defined over ..g'; 2.
(Vx)[~,cF"(:'(f2, x)] t":'] ~< [V~/Z(f2)] t":'l
T h e assertion is immediate for x -- 0. Assume as the induction hypothesis that the assertion is tue for all x < p. At x ~- p, if n~ = n p a , there is nothing to prove. So letn~ > n ~ l . Then k~ >~ k~_l , and hence, Fk,,/2(f2)~Fk(,-*)/2(f~). Also, by the induction hypothesis, t
t
t
t
[,~F~(')(f~., x)]t";-~] ~ [F~,,-',/~(A)] [";-,] ~< [F~'/2(f~)][n;J since F is well-behaved. By construction m(x) = k~/2 for n ~ l < x ~< n~', and thus
g(x) = F~/2(f2, x)
for
n'~_l ~< x ~ n~',
where
g(x) ---- [hxF~(~)(f2, x)].
Therefore, gin,'] ~< [Fko/2(f2)][~'] ' and the assertion is proved.
458
Basu
Consider now the definition of nx 9 It can be shown that the strict inequality could be weakened so that n, is the (/~z 9 aff)(Yfe jff~)[f[~] ~< [Fkd2(f2)][~] =~ Fk~'~(f, x) <~ Fk,(f2, x)] and therefore, (Yg 9 r162 ~< [Fk;2(f2)][";] * Fk'/~(g, x) <~Fk,(f~, x)] Since f l F-dominates f 2 , we have fx(X) > Fm)(f~., x) for almost of all x. Also, r(x) >~ k~ and hence, fl(X) > FY{X)(f2 , x) >~ Fk~(f2 , x) >/Fk;Z(g, x) ~ fl(x) > FlC'/Z(g, x) for almost all x. k~/2 is monotone, unbounded and thus fl F-dominates g. It is easy to verify that g F-dominates f~. Thus we havefl > r McFml~)(fz, x) >F f2 and the theorem is proved. Using theorem 2.1, we could effectively construct a chain of increasing recursive functions in the interval between the given functions f l and f~ such that the chain is densely ordered under the relation ofF-dominance. Consider now the well-behaved operator F obtained by the construction of L e m m a 1.2. F maps recursive functions into the set of recursive step-counting functions. Therefore, if f is recursive, F(f) is recursive and honest. Let re(x) be the function defined in Theorem 2.1. We then show the following, with the operator F as above: THEOREM 2.2. Let fx ,f2 be strictly increasing honest recursive functions and let r(x) be an elementary monotone recursive function such that r(x) < x and fl(x) > F~x)(f2 , x) for almost all x. Then AxV"lx)(f~ , x) is honest.
Proof. We consider F to be the well-behaved operator constructed in L e m m a 1.2. there is an O T M ~/'/'F that computes F(fz, x) using an f2 oracle Let g = )~xF"~)(f2, x), and let Tjd'~1')= g. .Ar is an O T M with f~ oracle that computes g(x) as follows. It computes m(x) Thus
by successively computing no, n a , n2 ,..., where ni is as defined in T h e o r e m 2.1, until it finds a y such that nu ~> x and sets m(x) = k~/z = a, say. Also by definition,
k~ = r(x) if r(x) is even = r(x) -- 1 if r(x)is odd. T h e n k~ is an elementary monotone function such that k~ < x. Let 6~ be the class of elementary functions../r then computes Fa(f2, x) using .//g,). It can be shown by induction on a that there is an r E do such that
qb~lPta x) ~ e(Fa(A x)) where aaxr
x) =
aaxF"(L, x). Thus for e' 9 do, ~ m ( x ) ~< e'(Fo(L, x) + ~,(x)),
where ~ = m. If we show that for some s ~ do, 9 ~(x) < s(g(x)),
(1)
SUBRECURSIVE DEGREES
459
then for s'r do, @~l~(x)< s'(g(x)). Let ~ ' , be a TM computing f2, so that (P,(x) < r .f~(x) for r ~ do since f2 is honest. Consider the TM ~'a obtained by putting together ~/r and ~ft'~I*) so that every time .//r questions its oracle as to whether x ~ T(f~), ~'v computes f2(xl) and compares with x0 where x = (xo, xl>. Now the longest argument at which ~ots*~ can interrogate its oracle and also the number of times it can interrogate the oracle are both bounded by ~12)(x) < s'(g(x)). Hence the total number of steps spent in computation of ,//~ is
< s'(g(x)),
max
O ~ y ~ s'(g(x))
q~(y)
-< s' (())o-<
since
~ , ( y ) < r .f~(y) < r .g(y)
s'(a(~))
< s'(g(x))
Z
r(gCi))
i---O
< tg(x)
for
t ~ do
Thus the number of steps needed by a standard TM (obtained by composing vr162and ~ft'~12)) to compute g(x) is bounded by t'(g(x)) for some t' ~ d~ which shows that g is honest. The rest of the proof is an extremely detailed analysis of the procedure for computing re(x), in order to prove assertion (1). The reader may prefer to skip this. Define F'(fx) = Ax[Fk,/~(f, x)] f o r f ~ ,A/'~r, k~ as before. Then F' is general recursive and is also well-behaved. For f ~ ,/ffw, let n J = (/zz ~ JV')(Vg ~ ./ffw)[f[~] = gt,] F'(f, x) = F'(g, x)]. Since F' is general recursive, there exists a recursive function a such that
F'(f, x) = ~ ( f (Zo), x) -- 1 where z o = n J = (t~z ~ ~/')[a(f(z), x) @ 0]. Since computing F'(f, x) involves computing n~t first, we have
n~I < F'(f, x). By definition, n~ is the length of the longest path in the finite subtree defining
{Fk,/~(f, x) If ~ 7J(h~)} where 7t(h~) = {f ~.A/"w I f < h~) and h~ = Fk,/~(f~). Therefore n~ < max n J f~Tt(hx)
< max Fk~/~(f, x) f~(h~)
(2)
For any f ~ ~ " and n e ~ff, let f~ be a recursive function that coincides with f on arguments up to n and is 0 elsewhere. Then f ~ 7t(h~) ~ ~ n ~ JV')(~f~ ~ ~ ) [ f t ~ ] = f i n ] andf~ ~< h~].
460 Setting
BASU
n
=
nJ , we have
FkJ2(f, x) = Fk'/2(f~ , x). Also,
:~ y~./'(f, x) <~ Y~.;'(h. , x). Hence, from (2),
n~ < max Fk'/2(f, x) ye~(h~)
< Fk,/2th,~, x) = F~,(A , x).
(3)
Consider now the tree defined by T(h~) = {f["] [ n e ..4/',f ~ r U { ~ }, ordered under set inclusion. (3) gives an upper bound on the height of the finite tree T~ defining {Fk,/2(f, x) IrE T(hx)}. The maximum degree at each vertex at level k is bounded by h k. Since f2 is strictly increasing and F is well-behaved, hk-x >/hk and hence the maximum degree in a tree of height n~ is bounded by h(n~). Let p(x) be the number of pendant vertices in T~. Then p(x) < 1--lin=%h(i). Let u(x) = Fk,/2(f2 , x), v(x) = Fk,(f2 , x), and w(x) ---- max(u(x), v(x)). Then f~ is increasing, F is well-behaved, and k~ is monotone, w(x) = v(x). Let r x) = w(x),
r
+ 1, x) = w(x)*(",~).
Then there exist nl, ng.E JV" such that u(x) < r
x) and v(x) < ~(n2, x). Therefore,
v(x)
P(x) < I-I uCi) <~ u(v(x)) ~'~' i=0
< r It is easy to show that there exists n ~ ~
p(x) < r
r
x))~("~'~).
such that
x) <~ e(w(x));
e~
= e(Fk'(f2, x)).
(4)
Consider now the computation of n, by a TM all,. Since n J < n~ for all f ~ ~b(h,) and n, < v(x), ./#n begins by computing u(0), u(1),..., u(v(x)) using the combination
SUBRECURSIVE DEGREES
461
JC'F of the machines ~r with anfz oracle and the T M .~r computingf. As shown earlier, the number of steps needed by .-r162is bounded by an elementary function of the number of steps needed by J/Cg~) withf~ oracle. Since k~ is elementary the number of steps spent by J[,~ in computing k~ may be ignored. Therefore, the number of steps needed to compute u(0),..., u(v(x))
< u(x) "uCv(x)) < r .vCx) for r e d~. Having computed u(x)t~t~], JC'n then computes n J for each f ~ W(h~). By the remark made above, it suffices to compute ftn~l
@,,(x) < <
~" r(v(x)) "u(x) + e(vCx)) y
+
e(v(x))
< s(v(x))
for some
= s(Fk*(f~, x)). We may choose s to be a strictly increasing elementary function, so that q~,, is bounded by an increasing function of x. The TM-/It now computes re(x) by successively computing n(0), n(1),..., till it finds a y such that nu ~> x and sets re(x) = k~/2. Then
re(x) @~(x) < E On(z) + r Z~O
where G = k. k is elementary implies @qis an elementary function, so we also have nx ~> x :~ y < x and thus m(x)= kJ2 < x. Therefore there is an elementary function t such that
t(x) < @~(y).
BASU
462 Therefore,
O~(x) < mCx) . ~,~ (m(x)) + t(x) x "O.(m(x)) + t(x) t'(O.CmCx))) + t(x)
(since re(x) < x; t' ~ #)
< t'(s(ek"~x'(f2 , re(x)))) + t(x)
s'(Fk~m~'(f2 , x)) + t(x)
(sincefz is increasing and re(x) < x)
< s'(F'~{x}(f~, x)) + t(x). We may choose f2 such that f2(x) > x so that for an elementary function s~,
9 ~(x) < s"(Fm{*)(f2 , x)). This proves the assertion and hence the theorem. COROLLARY 2.2. Let fl ,f~ be strictly increasing honest recursive functions and let fl >F f2 if there exists an elementary nondecreasing function r(x) such that r(x) < x and fl(x) > F~l~(f~ , x) for almost all x. Then there exists a dense chain of honest recursive functions ordered under >F between fl and f2.
Proof. Let fx ,f2 be strictly increasing honest recursive functions such that fx(X) > Frr 2 , x) for almost all x. In order to construct a dense chain we have to show that there exist elementary nondecreasing functions rl(x ) and r,(x) such that rl(x ) < x, r2(x ) < x, fl(x) > [Fr~{~[AxFm~x~(f2, x), x] and )txF'~x~(f2 , x) > F~J{~)(f~, x) for almost all x. From Theorem 2.1, fl(x) > F*;Z[AxFm'~'(f2 , x), x], where k~ = r(x) -- 1 = r(x)
if r(x) is odd, otherwise.
Let rx(X) = k~/2. Then rt(x ) < x, r I is elementary and nondecreasing. For a strictly increasing function f, define f-l(x) to be the least y such that f ( y ) ~ x if such a y exists, 0 otherwise. Then f-l(x) = x. Ritchie [4] has shown that i f f is honest then f-1 is a nondecreasing elementary function, and f - i f ( x ) ~ x. Further, i f f is any recursive function, then there is an honest, increasing recursive function h such that h(x) > f(x). Let g(x) be an honest strictly increasing function such that g(m(x)) > x, and let r2(x ) = g-l(x). Then r 2 is elementary, and r2(x ) < re(x) < x, and hence [AxFm{~}(f2, x)] (x) > F~2{~(f, x), which proves the corollary. The theorem proved above allows us to construct chains of honest recursive functions which are densely ordered under the F-dominance relation. Suppose now, f is any honest recursive function and let ~i = f . Then by definition, there is an elementary function r such that O~(x) < r 9~,(x). Let g be an honest strictly increasing recursive
SUBRECURSIVE DEGREES
463
function, majorizing all elementary functions such that g(x) > x for all x. Then, i f f is any honest recursive function, there is an index i such that ~i(x) < g ~(x). We now prove the main result of this section. LF.MMA2.3. Let F be a given complexity closure with respect to a general reeursive operator G. Then we can effectively construct a chain of honest recursive functions such that their closures under 1~, ordered under set inclusion, form a dense set.
Proof. Using the construction of Lemma 1.2, we can construct a well-behaved general recursive operator F' such that (VfE a~"r162[F'(f) > G(f)]. Define F"(f, x) = g(F'(f, x)), where g is as above. Clearly F" is general recursive. Applying Lemma 1.2 again, we can construct F such that (rye Jff~)[F(/) > F"(f)], and therefore,
F(f) > g(F'(f)). Since F' is well-behaved, F'(f) > f ~ F(f) > g 9f. Therefore, i f f = ~bi is any honest recursive function, F(f) > g "qbi > ~i. Let f l , f2, r(x) be as in Corollary 2.2 such that A(x) > F'C~(f2, x) for almost all x. Then
fl(x) > F~(f2, x)
for all k, and almost all x
fl(x) > (bI2(x) =~ A 9 F{A}
for almost all x, where r = f2.
by Axiom 3
:~ F{f2} C I'{A}
by Axiom 4.
Thus if ft >F f2, then the complexity closure of f~ is a subset of the complexity closure of f l . We next show that the inclusion is proper. By Axiom 3, we have oo
(Vg e F{f})(Vx)[F(f, x) > g(x)] since F ( f ) > G(f). Then for k > 0, Fk(f2) ~ F(f2}, ~o
for if it were, then by the above observation (Vx) F(f~, x) > Fk(f~, x), which is absurd since F is well-behaved. Let Cj = Fk(f2). Then Cj < F(r ----F(F~(f~)) = gk+l(f2). Also, by construction, Cj is honest. Since fl >F f2, we have fx(X) > Fk+l(f2 , x) for almost all x fl(x) > Cj(x) for almost all x Fk(A) ~ r{A}. Therefore for k > O, Fk(A) ~ P{f2} but Fk(f) 9 Hence, fl > r f2 :* F{A} D/'{f~}. This proves the lemma. Using Lemma 2.3, the main theorem is now immediate.
464
BASU
THEOREM 2.3. Let <~r be the reducibility ordering induced by a complexity closure F. Then there exist dense chains in the reducibility ordering of the I" degrees in the recursive Turning degree.
Proof. Letfx ,f2 be recursive functions such that F(f~) ~ F(fl). Then by Axiom 2 f2 ~ F(fl), and therefore f2 <~rfx. Also if fl E I'(f2), then by the same axiom /'(fx) _CF(f2) , contrary to hypothesis. Therefore, f2 ~ r f l but fi ~ r f ~ ; that is, f2 < Ffl 9 The theorem now follows from Lemma 2.3.
ACKNOWLEDGMENT The author wishes to thank Professor A. R. Meyer for suggesting the problem area and for several valuable discussions during the course of the work reported in this paper.
I~-'ERENCES 1. M. BLUM, A machine-independent theory of the complexity of recursive function, J. Assoc. Comput. Mach. 14 (1967), 323-336. 2. PAULAXT, On a subrecursive hierarchy and primitive recursive degrees, Trans. Amer. Math. Soc. 92 (1959), 85-105. 3. S. C. KLEENE, Extension of an effectively generated class of functions by enumeration, Colloq. Math. 6, 67-78. 4. D. M. RITCHIE, "Program Structure and Computational Complexity," Ph.D. Thesis, Harvard University, 1968. 5. H. ROGERS, JR., "Theory of Recursive Functions and Effective Computability," McGrawHill, New York, 1967. 6. S. K. BASU,On classes of computable functions, "Proceedings of the ACM Symposium on Theory of Computability," Los Angeles, May 1969.
(1970)
On the Structure of Subrecursive Degrees *t SANAT K. Basu:
Carnegie-Mellon University, Department of Computer Sciences, Pittsburgh, Penn. 15213 Received April 30, 1969
Subrecursive degrees are partitions of computable (recursive) functions generated by strong reducibility orderings. Such reducibilities can be naturally characterized in terms of closure operations. Closure operations corresponding to standard reducibilities such as "primitive recursive," etc., are computation time closed. It is shown that if the closure operation defining a strong reducibility satisfies certain axioms, then the partial ordering of the subrecursive degrees contains dense chains.
PRELIMINARIES Let X denote the nonnegative integers and ~ , ( ~ ) be the class of partial-recursive (recursive) functions of n variables. We shall use the standard indexing of partial recursive functions, satisfying the universal function theorem and the s-m-n theorem [1]. Let ~i denote the partial function computed by the i-th Turing Machine in the standard indexing. @i will denote the step counting function corresponding to ~i 9 @i(x) will be interpreted as the number of steps used by the i-th T.M. in computing ~i(x) if ~i(x) converges. ~i(x) is undefined otherwise. If ~i is a recursive function, then so is ~ i Axiomatic characterization of the step counting functions is given by Blum [1]. DEFINITION. A computable function f ( = ~i) is honest if the predicate [y = f(x)] is elementary in x and y. Equivalently, ~i is honest if there is an elementary function v(x, y) such that ~i(x) < v(x, $i(x)) for almost all x. A mapping F from functions of one variable to functions of one variable is called an operator. For any function f, the value of F(f) at x will be written as F(f, x). If 9 An earlier version of these results were presented at the ACM Symposium on Theory of Computing, Los Angeles, May 1969. 9 This work was supported by the Advanced Research Projects Agency of the Office of the Secretary of Defense (F 44620-67-C-0058) and is monitored by the Air Force Office of Scientific Research. 9 Present address: Computer Centre, Department of Electrical Engineering, Indian Institute of Technology, Kanpur, India. 452
SUBRECURSIVE DEGREES
453
f: ~V"~ ~/" let f(x) be an effective encoding o f f restricted to the domain {0, 1,..., x). Let .A/'~ be the class of total functions from ./V into ~ and let ~- be the class of all partial functions of one variable. DEFINITION. An operator F: ..UW--~ JV W is continuous if there is a function f of two arguments (called the associate of F) such that for every g e ,A/W, x e dff,
F(g, x) = f(~(z), x) -- l, where z is the least integer y such that
f(~(y), x) ~ O. DEFINITION. An operator F: .5"w--+ X W is general recursive if it is continuous and has an associate in ~2 9 I f f e J V W is a total function, then ft-I for n e~V" denotes the restriction o f f to the domain {0, 1,..., n}. LetF, g, and z be as above. Iff[zl ----gtZl thenF(g, x) = F(f, x), and further, z is the smallest integer for which this is true. Thus, if F is general recursive, then for all f ~ ./V"~r and all x (p.z ~ ~C) (Vg ~ ./VW) [f[~l = g[~] ~ F ( f , x) = F(g, x)] is defined and given the function f, can be computed using the recursive associate function of F.
I NTRODUCTION
The notion of primitive recursive degrees was first introduced by Kleene [3] and subsequently studied by Axt [2]. It was shown that the primitive recursive degrees in a given Turing degree form an upper semilattice and that there exist incomparable primitive recursive degrees in any given Turing degree. Restricting our attention to the recursive Turing degree, the primitive recursive degrees are still of interest, since they embody an interesting relation between computable functions, viz., the mutual ease of computability. If f and g are any two recursive functions, we say that f is primitive recursive in g if an oracle Turing Machine with a g oracle can computer in a number of steps bounded by a primitive recursive function. We say thatf and g belong to the same primitive recursive degree i f f is primitive recursive in g and g is primitive recursive in f. The choice of "primitive recursive" in the above discussion is clearly arbitrary. We could equally well consider "elementary degrees," "two-recursive degrees," etc. In general such reducibility orderings can be characterized by closure operations on functions such that a function f is reducible to a function g i f f is contained in the closure of g. All such closure operations must be effective if the induced reducibility ordering is to be stronger than Turing reducibility. Further they also have
454
BASU
the property of computation time closure, i.e., a function is in the class i f f f a (relative) step-counting function of it is also in the class. We have called the degrees of computable functions generated by reducibility ordering of the kind considered above "subrecursive degrees." Axt [2] has shown, that there exist denumerably many incomparble subrecursive degrees. Essentially similar results were obtained in Ref. [6] using only the boundedness property of the corresponding classes. In this paper we show that there exist dense chains in the reducibility ordering of the subrecursive degrees, using an axiomatization of the closure operations. In accordance with our preoccupation with computable functions, we have considered a closure to be a mapping, F, from sets of functions to sets of functions cO
satisfying the following axioms. Vx stands for almost all x. L e t / ' : 2 ~ X -~ 2 ~r'*" be a partial function./" is a complexity closure if there exists a general recursive operator F such that 1. f ~ ~1 ~ {f} ~ dom F
and
F{f} C ~x ;
2. f e r(g} ~ r(f} C r(g); o~
3. f e ~x =- (Vg e F(f})(Vx)[F(f, x) > g(x)]; 4. $j e ~x
and
f > r
~ $ i e F{f}.
We say that f is reducible to g i f f 6 F{g}. Axioms 1 and 2 are standard properties of closure. Axiom 3 provides an upper bound on the functions in the class and to that extent a closure satisfying Axiom 3 is constructive. Axiom 4 is weaker than the property of computation-time closure. In fact, if ~ is a closure operation such that q~(f) is computation-time closed a n d r e ~(f), then ~ satisfies Axiom 4. Clearly, therefore, standard closure operations such as "primitive recursive in," "elementary in" satisfy the axioms. Whether these are the weakest set of axioms under which the results stated in this paper can be derived is not known.
1. Well-Behaved Operators Consider a complexity closure F that satisfies Axioms 1--4. We now make precise the reducibility ordering induced b y / ' . DEFIr~ITION 1.1. A function f is F-reducible to a function g(f <~rg) if r E F(g). f is strictly F-reducible to g if f ~ r g and f ~ r g. By Axiom 2, ~
SUBRECURSIVE DEGREES
455
DEFINITION 1.2. Let G be any general recursive operator. The iterate of G is defined as
Go(f, x) = f(x), Ca(f, x) = C(f, x), Gn+l(f~ x) = G(Gr~(f), x). DEFINITION 1.3. Let G be a general recursive operator and f, g recursive functions. Then f G-dominates g(f >v g) if there exists a monotone, unbounded oo
recursive function r such that (Vx)[f(x) > Gr[z](g, x)]. DEFINITION 1.4. A general recursive operator G is well-behaved if the following hold for allf, g ~ J / ' ~ :
1. G(f) > f; 2. (Vx)[f(x) > g(x)] => (Vx)[G(f, x) > G(g, x)]; 3. (Vf e JV"z)(Vx, y)[x > y ~ G(f, x) > G(f,y)]; 4. (Vfe Mr~)(Vx 9 JV')(3 g 9 JV'~)[f[ x] -- g[=] and G(f, x) =/= G(g, x)]. Thus if is well-behaved, it preserves the majorising relation between functions. Also, G(f) is monotone for each total function f. (4) asserts that the (initial) segment o f f needed to compute G(f, x) is greater than x. Using induction it is easy to prove the following. LEMMA1.1. l f ageneral recursive operator G is well-behaved then so is Gn for all n ~ ~/*. We next prove the following, LEMMA 1.2. Let G be any given general recursive operator. Then we can effectively find a well-behaved general recursive operator F such that for all f ~ ~, F(f) > G(f)
Proof. Consider the step-counting function q~i of a Turing Machine computing a recursive functionf. Then r ~ x and r ~ f ( x ) , for firstly, the Turning Machine computing f(x) must read through the input before the computation could terminate, and secondly, the largest number the machine could write (in unary notation) is equal to the number of steps spent in computation. Suppose further that g 6 JV ~ is a total function and let ~r be a g oracle Turning Machine (OTM) with ~-(g) = ((x, g(x)) [ x ~ JV'} as the oracle, computing g. Then J/~g) needs at least g(x) steps to compute g(x) by sucessively questioning the oraele as to whether (x, O) ~ r(g), (x, 1) E -r(g),..., ( x, g(x)) E "r(g). Sinceg is total, this computation always halts. In terms of a relative step-counting function if r = g then ~b~~ g(x). For f E JV"~, let 7t(f, x) = {g ~ JV"W {g <~ Az max (f(z), x)}. Let gtq
456
BASU
be the restriction of g c ~g'~r to the domain {0, 1,..., n}. Let T(f, x) = {g[-l[ n c ~ - x , g e W(f, x)} U {;~}. Partially ordered under set inclusion, T(f, x) forms a tree with { ~} as the root and with finite branching at each vertex. Since G is general recursive and hence continuous, computation of G(g, x) for any g E ~(f, x) needs only a finite initial segment g. By K6nig's Lemma [5], for any x ~ JV', the set {G(g, x)l g ~ 7t(f, x)} is finite and is determined by a finite subset of T(f, x). In fact, S{G(g, x)i g E ~ ( f , x)}, as a function of x, is recursive in f. For f e ~ , let F(f, x) = q~1~(x), where ~b~11 -- h and
h(O) = [I{G(g, O) I g ~ 7t(f, 0)}] -t- f(0), h(x + 1) = [l{G(g,x + 1 ) [ g e T - ' ( f , x + 1)] + f ( x + 1) + F ( f , x ) . F is general recursive and maps recursive functions into the set of (recursive) stepcounting functions. Properties (1), (3), and (4) of well-behaved operators follow from the definition of F. We show that property (2) holds. Let f, k ~ .ArX, and (Vx)f(x) > k(x). Let h' be the function obtained by replacing f by k in the definition of h. Now (Vx)f(x) > k(x) ~ (Vx) tp(f, x) D ~(k, x) and also k 6 W(f, x). It follows then that F(f, 0) > F(k, 0). Assume as inductive hypothesis that F(f, x) > F(k, x). Then since the number of steps needed by a Turing Machine to compute a step-counting function is equal to the value of the function, and f(x) > k(x) => qb~"(x) > q~'(x) where ~ " - - f and ~k, = k, it is easily seen that F(f, x q- 1) > F(k, x + 1), which proves the lemma.
2. Dense Chain @Functions Suppose F is a given complexity closure with a general recursive operator G. Then using Lemma 1.2 above, we can effectively find a well-behaved general recursive operator F that also satisfies Axiom 3. Using this operator, we next prove the following: THEOREM 2.1. Let F be a well-behaved general recursive operator and let fx,f2 be strictly increasing recursive functions such that fl > ~fz. Then there exists a monotone, unbounded recursive function m such that
A >~ax[F'~'x'(A, x)] >~f~. Proof. fl > Ff2 implies that there exists a monotone, unbounded recursive function r' such thatfl(x ) >Fr'{x}(f2 , x)for almost all x. Since F ~ is well-behaved for all n ~ ~/', F"(f2) is also an increasing function of x. Let r(x) be a monotone unbounded recursive function such that r'(x) > r(x) for almost all x. Then fl(x) ), Fr{~l(f2, x) for almost all x. Define k~ = r(x) if r(x) is even, r(x) -- 1 otherwise. We shall construct m in stages, n~' denotes the segment o f f defined at the end of stage x.
457
SUBRECURSIVE DEGREES
Stage O. Set m(0) = 0, n o = 0. Stage x.
Compute kx 9 Let nx be the smallest integer z such that (Vfm,A/'~)[f ['] < [Fk'/e(f2)]tz] ~ F~'/2(f, x) < Fk'(f2 , x)] "-'.
(1)
We compute n~ as follows: Let hx :Fk'/~(f2). Then h , ~ , and let T(h,) : { f e J f f ' ~ I f < h~}. Define T(hx) = {fin] I n ~ h / ' , f e 7t(h,)} u {~}. By an argument parallel to that used in the proof of Lemma 1.2, there is a finite subset of T(h~) that defines the set {Fk,/2(f, x ) l y e T(h,)}. This finite subset corresponds to a finite tree obtained from T(h,) by ordering it under set inclusion. Let n, be the length of the longest path in this subtree. SinceF is general recursive and h~ is recursive, n~ is effectively computable. Then for all g ~ JV"~', gtn,] < [Fk:12(f2)][,,,] ~ (3 g' E Jff~)[g' E tP(h~) and g'[",] :
gin=I]. Also, g ' e 7J(h~)=~ g ' <
h~ and since F is well-behaved, Fk,/2(g ')
Fk'(f2). Further,
by
construction
of
n~, n~ ~ (~z c ~U)(Vf c jCr~r)[f[z] _-- g,[~] =v
tk,/2(f, x) = p,/2(g,, x)]. But g[n,] = g,[n,], which implies that Fk,/2(g, x) = Fk,/2(g ', x). Then for all g e .A/"~, gln,] .< [Fk:/~(A)][n,] ~ F~:12(g, x) =- Fk:/2(g ', x) .< Fk,12(F~,/2(f~ , x) = F~,(f2 , x). Therefore n~ has the property asserted in (1). Let n~' = max[n~, n~_i], n o' - 0. Set re(y) = k~/2 for n~_ 1 ~< y ~< n~. We next show #
1.
#
n~' is an unbounded function of x.
By construction of n~, for any g ~ 7x(P:/2(f2)), n, >~ (t~z ~ ./V')(Vf~ .,e'~)[f[:] = g[:l =~ p : / 2 ( f , x) = F~:12(g, x)]. F ~,/~ is well-behaved for all x, and therefore by condition 4 of Definition 1.4, n~ ~ x. Thus n~' is unbounded and m gets defined over ..g'; 2.
(Vx)[~,cF"(:'(f2, x)] t":'] ~< [V~/Z(f2)] t":'l
T h e assertion is immediate for x -- 0. Assume as the induction hypothesis that the assertion is tue for all x < p. At x ~- p, if n~ = n p a , there is nothing to prove. So letn~ > n ~ l . Then k~ >~ k~_l , and hence, Fk,,/2(f2)~Fk(,-*)/2(f~). Also, by the induction hypothesis, t
t
t
t
[,~F~(')(f~., x)]t";-~] ~ [F~,,-',/~(A)] [";-,] ~< [F~'/2(f~)][n;J since F is well-behaved. By construction m(x) = k~/2 for n ~ l < x ~< n~', and thus
g(x) = F~/2(f2, x)
for
n'~_l ~< x ~ n~',
where
g(x) ---- [hxF~(~)(f2, x)].
Therefore, gin,'] ~< [Fko/2(f2)][~'] ' and the assertion is proved.
458
Basu
Consider now the definition of nx 9 It can be shown that the strict inequality could be weakened so that n, is the (/~z 9 aff)(Yfe jff~)[f[~] ~< [Fkd2(f2)][~] =~ Fk~'~(f, x) <~ Fk,(f2, x)] and therefore, (Yg 9 r162 ~< [Fk;2(f2)][";] * Fk'/~(g, x) <~Fk,(f~, x)] Since f l F-dominates f 2 , we have fx(X) > Fm)(f~., x) for almost of all x. Also, r(x) >~ k~ and hence, fl(X) > FY{X)(f2 , x) >~ Fk~(f2 , x) >/Fk;Z(g, x) ~ fl(x) > FlC'/Z(g, x) for almost all x. k~/2 is monotone, unbounded and thus fl F-dominates g. It is easy to verify that g F-dominates f~. Thus we havefl > r McFml~)(fz, x) >F f2 and the theorem is proved. Using theorem 2.1, we could effectively construct a chain of increasing recursive functions in the interval between the given functions f l and f~ such that the chain is densely ordered under the relation ofF-dominance. Consider now the well-behaved operator F obtained by the construction of L e m m a 1.2. F maps recursive functions into the set of recursive step-counting functions. Therefore, if f is recursive, F(f) is recursive and honest. Let re(x) be the function defined in Theorem 2.1. We then show the following, with the operator F as above: THEOREM 2.2. Let fx ,f2 be strictly increasing honest recursive functions and let r(x) be an elementary monotone recursive function such that r(x) < x and fl(x) > F~x)(f2 , x) for almost all x. Then AxV"lx)(f~ , x) is honest.
Proof. We consider F to be the well-behaved operator constructed in L e m m a 1.2. there is an O T M ~/'/'F that computes F(fz, x) using an f2 oracle Let g = )~xF"~)(f2, x), and let Tjd'~1')= g. .Ar is an O T M with f~ oracle that computes g(x) as follows. It computes m(x) Thus
by successively computing no, n a , n2 ,..., where ni is as defined in T h e o r e m 2.1, until it finds a y such that nu ~> x and sets m(x) = k~/z = a, say. Also by definition,
k~ = r(x) if r(x) is even = r(x) -- 1 if r(x)is odd. T h e n k~ is an elementary monotone function such that k~ < x. Let 6~ be the class of elementary functions../r then computes Fa(f2, x) using .//g,). It can be shown by induction on a that there is an r E do such that
qb~lPta x) ~ e(Fa(A x)) where aaxr
x) =
aaxF"(L, x). Thus for e' 9 do, ~ m ( x ) ~< e'(Fo(L, x) + ~,(x)),
where ~ = m. If we show that for some s ~ do, 9 ~(x) < s(g(x)),
(1)
SUBRECURSIVE DEGREES
459
then for s'r do, @~l~(x)< s'(g(x)). Let ~ ' , be a TM computing f2, so that (P,(x) < r .f~(x) for r ~ do since f2 is honest. Consider the TM ~'a obtained by putting together ~/r and ~ft'~I*) so that every time .//r questions its oracle as to whether x ~ T(f~), ~'v computes f2(xl) and compares with x0 where x = (xo, xl>. Now the longest argument at which ~ots*~ can interrogate its oracle and also the number of times it can interrogate the oracle are both bounded by ~12)(x) < s'(g(x)). Hence the total number of steps spent in computation of ,//~ is
< s'(g(x)),
max
O ~ y ~ s'(g(x))
q~(y)
-< s' (())o-<
since
~ , ( y ) < r .f~(y) < r .g(y)
s'(a(~))
< s'(g(x))
Z
r(gCi))
i---O
< tg(x)
for
t ~ do
Thus the number of steps needed by a standard TM (obtained by composing vr162and ~ft'~12)) to compute g(x) is bounded by t'(g(x)) for some t' ~ d~ which shows that g is honest. The rest of the proof is an extremely detailed analysis of the procedure for computing re(x), in order to prove assertion (1). The reader may prefer to skip this. Define F'(fx) = Ax[Fk,/~(f, x)] f o r f ~ ,A/'~r, k~ as before. Then F' is general recursive and is also well-behaved. For f ~ ,/ffw, let n J = (/zz ~ JV')(Vg ~ ./ffw)[f[~] = gt,] F'(f, x) = F'(g, x)]. Since F' is general recursive, there exists a recursive function a such that
F'(f, x) = ~ ( f (Zo), x) -- 1 where z o = n J = (t~z ~ ~/')[a(f(z), x) @ 0]. Since computing F'(f, x) involves computing n~t first, we have
n~I < F'(f, x). By definition, n~ is the length of the longest path in the finite subtree defining
{Fk,/~(f, x) If ~ 7J(h~)} where 7t(h~) = {f ~.A/"w I f < h~) and h~ = Fk,/~(f~). Therefore n~ < max n J f~Tt(hx)
< max Fk~/~(f, x) f~(h~)
(2)
For any f ~ ~ " and n e ~ff, let f~ be a recursive function that coincides with f on arguments up to n and is 0 elsewhere. Then f ~ 7t(h~) ~ ~ n ~ JV')(~f~ ~ ~ ) [ f t ~ ] = f i n ] andf~ ~< h~].
460 Setting
BASU
n
=
nJ , we have
FkJ2(f, x) = Fk'/2(f~ , x). Also,
:~ y~./'(f, x) <~ Y~.;'(h. , x). Hence, from (2),
n~ < max Fk'/2(f, x) ye~(h~)
< Fk,/2th,~, x) = F~,(A , x).
(3)
Consider now the tree defined by T(h~) = {f["] [ n e ..4/',f ~ r U { ~ }, ordered under set inclusion. (3) gives an upper bound on the height of the finite tree T~ defining {Fk,/2(f, x) IrE T(hx)}. The maximum degree at each vertex at level k is bounded by h k. Since f2 is strictly increasing and F is well-behaved, hk-x >/hk and hence the maximum degree in a tree of height n~ is bounded by h(n~). Let p(x) be the number of pendant vertices in T~. Then p(x) < 1--lin=%h(i). Let u(x) = Fk,/2(f2 , x), v(x) = Fk,(f2 , x), and w(x) ---- max(u(x), v(x)). Then f~ is increasing, F is well-behaved, and k~ is monotone, w(x) = v(x). Let r x) = w(x),
r
+ 1, x) = w(x)*(",~).
Then there exist nl, ng.E JV" such that u(x) < r
x) and v(x) < ~(n2, x). Therefore,
v(x)
P(x) < I-I uCi) <~ u(v(x)) ~'~' i=0
< r It is easy to show that there exists n ~ ~
p(x) < r
r
x))~("~'~).
such that
x) <~ e(w(x));
e~
= e(Fk'(f2, x)).
(4)
Consider now the computation of n, by a TM all,. Since n J < n~ for all f ~ ~b(h,) and n, < v(x), ./#n begins by computing u(0), u(1),..., u(v(x)) using the combination
SUBRECURSIVE DEGREES
461
JC'F of the machines ~r with anfz oracle and the T M .~r computingf. As shown earlier, the number of steps needed by .-r162is bounded by an elementary function of the number of steps needed by J/Cg~) withf~ oracle. Since k~ is elementary the number of steps spent by J[,~ in computing k~ may be ignored. Therefore, the number of steps needed to compute u(0),..., u(v(x))
< u(x) "uCv(x)) < r .vCx) for r e d~. Having computed u(x)t~t~], JC'n then computes n J for each f ~ W(h~). By the remark made above, it suffices to compute ftn~l
@,,(x) < <
~" r(v(x)) "u(x) + e(vCx)) y
+
e(v(x))
< s(v(x))
for some
= s(Fk*(f~, x)). We may choose s to be a strictly increasing elementary function, so that q~,, is bounded by an increasing function of x. The TM-/It now computes re(x) by successively computing n(0), n(1),..., till it finds a y such that nu ~> x and sets re(x) = k~/2. Then
re(x) @~(x) < E On(z) + r Z~O
where G = k. k is elementary implies @qis an elementary function, so we also have nx ~> x :~ y < x and thus m(x)= kJ2 < x. Therefore there is an elementary function t such that
t(x) < @~(y).
BASU
462 Therefore,
O~(x) < mCx) . ~,~ (m(x)) + t(x) x "O.(m(x)) + t(x) t'(O.CmCx))) + t(x)
(since re(x) < x; t' ~ #)
< t'(s(ek"~x'(f2 , re(x)))) + t(x)
s'(Fk~m~'(f2 , x)) + t(x)
(sincefz is increasing and re(x) < x)
< s'(F'~{x}(f~, x)) + t(x). We may choose f2 such that f2(x) > x so that for an elementary function s~,
9 ~(x) < s"(Fm{*)(f2 , x)). This proves the assertion and hence the theorem. COROLLARY 2.2. Let fl ,f~ be strictly increasing honest recursive functions and let fl >F f2 if there exists an elementary nondecreasing function r(x) such that r(x) < x and fl(x) > F~l~(f~ , x) for almost all x. Then there exists a dense chain of honest recursive functions ordered under >F between fl and f2.
Proof. Let fx ,f2 be strictly increasing honest recursive functions such that fx(X) > Frr 2 , x) for almost all x. In order to construct a dense chain we have to show that there exist elementary nondecreasing functions rl(x ) and r,(x) such that rl(x ) < x, r2(x ) < x, fl(x) > [Fr~{~[AxFm~x~(f2, x), x] and )txF'~x~(f2 , x) > F~J{~)(f~, x) for almost all x. From Theorem 2.1, fl(x) > F*;Z[AxFm'~'(f2 , x), x], where k~ = r(x) -- 1 = r(x)
if r(x) is odd, otherwise.
Let rx(X) = k~/2. Then rt(x ) < x, r I is elementary and nondecreasing. For a strictly increasing function f, define f-l(x) to be the least y such that f ( y ) ~ x if such a y exists, 0 otherwise. Then f-l(x) = x. Ritchie [4] has shown that i f f is honest then f-1 is a nondecreasing elementary function, and f - i f ( x ) ~ x. Further, i f f is any recursive function, then there is an honest, increasing recursive function h such that h(x) > f(x). Let g(x) be an honest strictly increasing function such that g(m(x)) > x, and let r2(x ) = g-l(x). Then r 2 is elementary, and r2(x ) < re(x) < x, and hence [AxFm{~}(f2, x)] (x) > F~2{~(f, x), which proves the corollary. The theorem proved above allows us to construct chains of honest recursive functions which are densely ordered under the F-dominance relation. Suppose now, f is any honest recursive function and let ~i = f . Then by definition, there is an elementary function r such that O~(x) < r 9~,(x). Let g be an honest strictly increasing recursive
SUBRECURSIVE DEGREES
463
function, majorizing all elementary functions such that g(x) > x for all x. Then, i f f is any honest recursive function, there is an index i such that ~i(x) < g ~(x). We now prove the main result of this section. LF.MMA2.3. Let F be a given complexity closure with respect to a general reeursive operator G. Then we can effectively construct a chain of honest recursive functions such that their closures under 1~, ordered under set inclusion, form a dense set.
Proof. Using the construction of Lemma 1.2, we can construct a well-behaved general recursive operator F' such that (VfE a~"r162[F'(f) > G(f)]. Define F"(f, x) = g(F'(f, x)), where g is as above. Clearly F" is general recursive. Applying Lemma 1.2 again, we can construct F such that (rye Jff~)[F(/) > F"(f)], and therefore,
F(f) > g(F'(f)). Since F' is well-behaved, F'(f) > f ~ F(f) > g 9f. Therefore, i f f = ~bi is any honest recursive function, F(f) > g "qbi > ~i. Let f l , f2, r(x) be as in Corollary 2.2 such that A(x) > F'C~(f2, x) for almost all x. Then
fl(x) > F~(f2, x)
for all k, and almost all x
fl(x) > (bI2(x) =~ A 9 F{A}
for almost all x, where r = f2.
by Axiom 3
:~ F{f2} C I'{A}
by Axiom 4.
Thus if ft >F f2, then the complexity closure of f~ is a subset of the complexity closure of f l . We next show that the inclusion is proper. By Axiom 3, we have oo
(Vg e F{f})(Vx)[F(f, x) > g(x)] since F ( f ) > G(f). Then for k > 0, Fk(f2) ~ F(f2}, ~o
for if it were, then by the above observation (Vx) F(f~, x) > Fk(f~, x), which is absurd since F is well-behaved. Let Cj = Fk(f2). Then Cj < F(r ----F(F~(f~)) = gk+l(f2). Also, by construction, Cj is honest. Since fl >F f2, we have fx(X) > Fk+l(f2 , x) for almost all x fl(x) > Cj(x) for almost all x Fk(A) ~ r{A}. Therefore for k > O, Fk(A) ~ P{f2} but Fk(f) 9 Hence, fl > r f2 :* F{A} D/'{f~}. This proves the lemma. Using Lemma 2.3, the main theorem is now immediate.
464
BASU
THEOREM 2.3. Let <~r be the reducibility ordering induced by a complexity closure F. Then there exist dense chains in the reducibility ordering of the I" degrees in the recursive Turning degree.
Proof. Letfx ,f2 be recursive functions such that F(f~) ~ F(fl). Then by Axiom 2 f2 ~ F(fl), and therefore f2 <~rfx. Also if fl E I'(f2), then by the same axiom /'(fx) _CF(f2) , contrary to hypothesis. Therefore, f2 ~ r f l but fi ~ r f ~ ; that is, f2 < Ffl 9 The theorem now follows from Lemma 2.3.
ACKNOWLEDGMENT The author wishes to thank Professor A. R. Meyer for suggesting the problem area and for several valuable discussions during the course of the work reported in this paper.
I~-'ERENCES 1. M. BLUM, A machine-independent theory of the complexity of recursive function, J. Assoc. Comput. Mach. 14 (1967), 323-336. 2. PAULAXT, On a subrecursive hierarchy and primitive recursive degrees, Trans. Amer. Math. Soc. 92 (1959), 85-105. 3. S. C. KLEENE, Extension of an effectively generated class of functions by enumeration, Colloq. Math. 6, 67-78. 4. D. M. RITCHIE, "Program Structure and Computational Complexity," Ph.D. Thesis, Harvard University, 1968. 5. H. ROGERS, JR., "Theory of Recursive Functions and Effective Computability," McGrawHill, New York, 1967. 6. S. K. BASU,On classes of computable functions, "Proceedings of the ACM Symposium on Theory of Computability," Los Angeles, May 1969.